Determine the value ofIntegral from negative 1 to 1 StartFraction 1 Over x EndFraction dx111xdx.Question content area bottomPart 1Choose the correct answer below.A.Integral from negative 1 to 1 StartFraction 1 Over x EndFraction dx111xdxdiverges becauseModifyingBelow lim With t right arrow 0 Superscript pluslimt0+ln left parenthesis t right parenthesisln(t)does not exist.B.Integral from negative 1 to 1 StartFraction 1 Over x EndFraction dx111xdxequals=0becauseStartFraction 1 Over x EndFraction1xis odd.C.Integral from negative 1 to 1 StartFraction 1 Over x EndFraction dx111xdxdiverges becauseln left parenthesis negative 1 right parenthesisln(1)is not defined.D.Integral from negative 1 to 1 StartFraction 1 Over x EndFraction dx equals ModifyingBelow lim With b right arrow 0 Superscript minus Integral from negative 1 to b StartFraction 1 Over x EndFraction dx plus ModifyingBelow lim With b right arrow 0 Superscript plus Integral from b to 1 StartFraction 1 Over x EndFraction dx equals ln 2Integral from negative 1 to 1 StartFraction 1 Over x EndFraction dx111xdxequals=ModifyingBelow lim With b right arrow 0 Superscript minuslimb0Integral from negative 1 to b StartFraction 1 Over x EndFraction dx1b1xdxplus+ModifyingBelow lim With b right arrow 0 Superscript pluslimb0+Integral from b to 1 StartFraction 1 Over x EndFraction dxb11xdxequals=ln 2